Solution:
The small signal drain current of $M_1=g_{m_1}V_{in}$
Also $I_{D_1}=I_{D_2} \quad \therefore I_{D_2}=g_{m_1}V_{in}$
$M_2\ \&\ M_3$ forms a basic current mirror
$As,\ I_{D_3}=\frac{{(\frac{W}{L})}_3 }{{(\frac{W}{L})}_2} I_{D_2}$
$\therefore I_{D_3}=\frac{{(\frac{W}{L})}_3 }{{(\frac{W}{L})}_2}g_{m_1}V_{in}$
$As,\ V_{out}=R_L I_{D_3}$
$\therefore V_{out}=R_L \frac{{(\frac{W}{L})}_3 }{{(\frac{W}{L})}_2}g_{m_1}V_{in}$
$\therefore A_v=\frac{V_{out}}{V_{in}}=R_L \frac{{(\frac{W}{L})}_3 }{{(\frac{W}{L})}_2}g_{m_1}$